Question

Is series
$\sqrt{3}$

$\sqrt{6}$

$\sqrt{9}$
$\sqrt{12}$
.......an AP ? give reason​

Answer 1

Let's find whether there is a common difference between each consecutive terms or not.

There's no need to find the difference between all. Just find the differences between first and second terms, and then second and third, and check whether they're same or not.

$\sqrt{6}-\sqrt{3} \\ \\ \sqrt{2\times3} - \sqrt{3} \\ \\ \sqrt{2} \times \sqrt{3} - \sqrt{3} \\ \\ \sqrt{3}(\sqrt{2}-1) \\ \\ \\ \sqrt{9}-\sqrt{6} \\ \\ \sqrt{3\times3} - \sqrt{2\times3} \\ \\ \sqrt{3} \times \sqrt{3} - \sqrt{2} \times \sqrt{3} \\ \\ \sqrt{3}(\sqrt{3}-\sqrt{2})$

Here it doesn't seem that both are equal.

$\sqrt{6}-\sqrt{3} \neq \sqrt{9}-\sqrt{6}$

So it's not an AP.

Hope my answer may help you. ^_^

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Thank you. :-))